Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
// This template code suggested by KT BYTE Computer Science Academy
//   for use in reading and writing files for USACO problems.

// https://content.ktbyte.com/problem.java

import java.util.*;
import java.io.*;

public class Main {

    static StreamTokenizer in;

    static int nextInt() throws IOException {
        in.nextToken();
        return (int) in.nval;
    }

    static double nextDouble() throws IOException {
        in.nextToken();
        return in.nval;
    }

    public static void main(String[] args) throws Exception {
        in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));

        int n = nextInt();

        double[] probabilities = new double[n];

        for (int i = 0; i < n; i++) {
            probabilities[i] = nextDouble();
        }

        double[][] dp = new double[n + 1][n + 1]; // coins, and the number of heads = probabilities

        dp[0][0] = 1.0;

        for (int i = 1; i <= n; i++) {
            for (int j = 0; j <= i; j++) {
                if (j - 1 >= 0) dp[i][j] = dp[i-1][j-1] * probabilities[i - 1];
                dp[i][j] += dp[i-1][j] * (1 - probabilities[i - 1]);
            }
        }

        double result = 0;

        for (int j = 0; j <= n; j++) {
            if (j > n - j) {
                result += dp[n][j];
            }
        }

        out.println(result);

        out.close();
    }
}